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![](https://rs.olm.vn/images/avt/0.png?1311)
Xét vế trái :\(3^{-1}.3^n+6.3^{n-1}=\frac{1}{3}.3^n+6.3^{n-1}=3^{n-1}+6.3^{n-1}=7.3^{n-1}\)
So sánh với vế phải , suy ra \(3^{n-1}=3^6\Leftrightarrow n-1=6\Leftrightarrow n=7\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt A=\(\dfrac{1}{1.2.3}\)+\(\dfrac{1}{2.3.4}\)+\(\dfrac{1}{3.4.5}\)+...+\(\dfrac{1}{n\left(n+1\right)\left(n+2\right)}\)
=>2A=\(\dfrac{2}{1.2.3}\)+\(\dfrac{2}{2.3.4}\)+...+\(\dfrac{2}{n\left(n+1\right)\left(n+2\right)}\)
=\(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\)\(\dfrac{1}{n\left(n+1\right)}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\)
=\(\dfrac{1}{2}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\)
=\(\dfrac{\left(n+1\right)\left(n+2\right)-2}{2\left(n+1\right)\left(n+2\right)}\)
=\(\dfrac{n^2+3n}{2\left(n^2+3n+2\right)}\)
=>A=\(\dfrac{n^2+3n}{4n^2+12n+8}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=1.2.3+2.3.4+...+n\left(n+1\right)\left(n+2\right)\)
\(4S=1.2.3.4+2.3.4.4+...+n\left(n+1\right)\left(n+2\right).4\)
\(4S=1.2.3.4+2.3.4.\left(5-1\right)+...+n\left(n+1\right)\left(n+2\right)\)
\(\left[\left(n+3\right)-\left(n-1\right)\right]\)
\(4S=1.2.3.4+2.3.4.5-1.2.3.4+...+\)
\(n\left(n+1\right)\left(n+2\right)\left(n+3\right)-\left(n-1\right)n\left(n+1\right)\left(n+2\right)\)
\(4S=n\left(n+1\right)\left(n+2\right)\left(n+3\right)\)
\(4S+1=n\left(n+3\right)\left(n+1\right)\left(n+2\right)+1\)
\(=\left(n^2+3n\right)\left(n^2+3n+2\right)+1\)
Đặt \(n^2+3n=t\)
\(Đt=t\left(t+2\right)+1=t^2+2t+1=\left(t+1\right)^2\)(là số chính phương)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=\frac{5}{1.2.3}+\frac{5}{2.3.4}+...+\frac{5}{n.\left(n+1\right)\left(n+2\right)}\)
\(\Leftrightarrow\frac{2B}{5}=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{n\left(n+1\right)\left(n+2\right)}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)
\(=\frac{1}{2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)
\(\Rightarrow B=\frac{5}{4}-\frac{5}{2\left(n+1\right)\left(n+2\right)}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a^2 + b^2 + c^2= ab + bc + ca
2 ( a^2 + b^2 + c^2 ) = 2 ( ab + bc + ca)
2a^2 + 2b^2 + 2c^2 = 2ab + 2bc + 2ca
a^2 + a^2 + b^2 + b^2 + c^2+ c^2 – 2ab – 2bc – 2ca = 0
a^2 + b^2 – 2ab + b^2 + c^2 – 2bc + c² + a² – 2ca = 0
(a^2 + b^2 – 2ab) + (b^2 + c^2 – 2bc) + (c^2 + a^2 – 2ca) = 0
(a – b)^2 + (b – c)^2 + (c – a)^2 = 0
Vì (a-b)^2 lớn hơn hoặc bằng 0 với mọi a và b
(b-c)^2 lớn hơn hoặc bằng 0 với mọi c và b
(c-a)^2 lớn hơn hoặc bằng 0 với mọi a và c
=> (a-b)^2 =0 ; (b-c)^2=0 ; (c-a)^2=0
=> a=b ; b=c ; c=a
=>a=b=c
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{2}\left(\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)\)
\(\Rightarrow\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{2005.2006.2007}=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+...+\frac{1}{2005.2006}-\frac{1}{2006.2007}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2006.2007}\right)=\frac{1}{2}\left(\frac{2005.2008}{2.2006.2007}\right)\)
Đặt \(A=1.2+2.3+...+n\left(n+1\right)\)
\(\Rightarrow3A=1.2.\left(3-0\right)+2.3.\left(4-1\right)+...+n\left(n+1\right)\left(n+2-\left(n-1\right)\right)\)
\(\Rightarrow3A=1.2.3-1.2.0+2.3.4-1.2.3+...+n\left(n+1\right)\left(n+2\right)-\left(n-1\right)n\left(n+1\right)\)
\(\Rightarrow3A=n\left(n+1\right)\left(n+2\right)\)
\(\Rightarrow A=\frac{n\left(n+1\right)\left(n+2\right)}{3}\)
\(\Rightarrow1.2+2.3+...+2006.2007=\frac{2006.2007.2008}{2}\)
Vậy pt trở thành:
\(\frac{1}{2}\left(\frac{2005.2008}{2.2006.2007}\right)x=\frac{2006.2007.2008}{2}\)
\(\Leftrightarrow\frac{2005}{2.2006.2007}x=2006.2007\)
\(\Rightarrow x=\frac{2.\left(2006.2007\right)^2}{2005}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
B=1/2.1.2-1/2.2.3+1/2.2.3-1/2.3.4+...+1/2n(n+1)-1/2(n+1)(n+2)
B=1/2[(1/1.2+1/2.3+...+1/n(n+1))-(1/2.3+1/3.4+...+1/(n+1)(n+2))]
Tới đây bạn tự làm tiếp nha, tương tự như bài 1/1.2+1/2.3+..+1/n(n+1) á bạn.Cái này bạn ghi ra bạn sẽ hiểu, mình viết hơi bị lủng củng.
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
Ta có: \(\frac{1}{k(k+1)(k+2)}=\frac{1}{2}.\frac{2}{k(k+1)(k+2)}=\frac{1}{2}.\frac{(k+2)-k}{k(k+1)(k+2)}\)
\(=\frac{1}{2}\left(\frac{k+2}{k(k+1)(k+2)}-\frac{k}{k(k+1)(k+2)}\right)=\frac{1}{2}\left(\frac{1}{k(k+1)}-\frac{1}{(k+1)(k+2)}\right)\)
Áp dụng vào bài toán:
\(\frac{1}{1.2.3}=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}\right)\)
\(\frac{1}{2.3.4}=\frac{1}{2}\left(\frac{1}{2.3}-\frac{1}{3.4}\right)\)
\(\frac{1}{3.4.5}=\frac{1}{2}\left(\frac{1}{3.4}-\frac{1}{4.5}\right)\)
.......
\(\frac{1}{n(n+1)(n+2)}=\frac{1}{2}\left(\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)}\right)\)
\(\Rightarrow B=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{(n+1)(n+2)}\right)=\frac{1}{4}-\frac{1}{2(n+1)(n+2)}\)
vào đây nha https://coccoc.com/search/math#query=+3%5E%E2%88%921%C2%B73%5En%2B6%C2%B73%5En%E2%88%921%3D7%C2%B736++
\(3^{-1}\cdot3^n+6\cdot3^{n-1}=7\cdot3^6\)
\(3^{n-1}+6\cdot3^{n-1}=7\cdot3^6\)
\(3^{n-1}\left(1+6\right)=7\cdot3^6\)
\(3^{n-1}\cdot7=7\cdot3^6\)
\(\Rightarrow3^{n-1}=3^6\)
\(\Rightarrow n-1=6\)
\(n=6+1=7\)